12 Examples The two maximal executions of a b, which are actually equal. If a = b, one more execution. An execution of a µx.(x X ). Some wrong examples. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 12/38

13 Restrictions and moves A restriction from X to Y is a cospan Y X id X. A move from X to Y is a cospan Y M X of presheaves obtained t s s from a cospan Y 0 M 0 X 0, with M 0 a representable of dim 2 or 3, by identifying some names. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 13/38

16 A Grothendieck topology Let have dimension 0, n have dimension 1, and so on up to 3. Let a sieve S on X U in E be view-covering when it is jointly surjective in dimension 1. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 16/38

17 Elementary views An elementary view from X to Y is a composite of a move from a representable, followed by a restriction to a representable: n X Keeps track of one trajectory. id X M n. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 17/38

18 Views A view is an observation X U isomorphic to a possibly denumerable composition of elementary views. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 18/38

19 Views are a canonical covering Proposition For any observation X U, the sieve generated by morphisms from finite views into U is covering. Proposition Any covering sieve contains all morphisms from finite views. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 19/38

20 The categories E X Here we want to relativise to a base position X. Let E X have as objects U Y X, and morphisms transformations between such with X fixed. U U Y Y E X inherits a Grothendieck topology from E. X Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 20/38

22 The stack of strategies Proposition This X S X extends to a functor S: B op CAT, which is a stack for the restriction of the view-covering topology to B. Why stacks? Strategies are only sensible up to iso. Intuitively, only the number of possible states should matter, not the precise set of states. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 22/38

24 Temporal decomposition Let M X be the set of moves from X (explain the size). For each i M X, let X i be the domain of i. Theorem S n Fam i M n S Xi. A strategy is determined by its initial states, and what remains of them after each possible move. Almost a sketch: would be S n = i M n S Xi. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 24/38

26 Must testing Supposing a fixed move : A process P is must orthogonal to a context C, when all maximal traces of C[P] play at some point. Notation: P m C, P m. P and Q are must equivalent, notation P m Q, when P m = Q m. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 26/38

27 Must testing in an unfair setting Usually, only the unfair scenario is formalised: P = (Ω a) and Q = Ω are must equivalent. The obvious test C = a. is not orthogonal to P. Indeed, there is an infinite looping trace, maximal. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 27/38

28 Fair testing in an unfair setting The example (Ω a) m Ω takes potential unfairness of the scheduler into account. Usually people do not want to, and resort to: A process P is fair orthogonal to a context C, when all finite traces of C[P] extend to traces that play at some point. Notation: P f C, P f. P and Q are fair equivalent, notation P f Q, when P f = Q f. Solves the issue. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 28/38

32 Interactive equivalence For any strategy S on X and any pushout P I X Y Z (2) of positions with I of dimension 0, let S P be the class of all strategies T on Y such that Gl(S T ) Z. Here denotes amalgamation in the stack S. Let us make this concrete. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 32/38

33 Fair testing A closed-world story is successful when it contains a n. Given a global behaviour G G X, an extension of a state s G(U) to U is an s G(U ) with i : U U and s i = s. The fair criterion f X contains all global behaviours G such that any state s G(U) for finite U admits a successful extension. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 33/38

34 Must testing An extension of s G(U) is strict when U U is not surjective. For any global behaviour G G X, a state s G(U) is G-maximal when it has no strict extension. Let the must criterion m X consist of all global behaviours G such that for all closed-world U, and G-maximal s G(U), U is successful. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 34/38

35 The key result Theorem For any strategy S, any state s Gl(S)(U) admits a Gl(S)-maximal extension. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 35/38

36 Fair vs. must Thanks to the theorem, we have: Lemma For all S S X, Gl(S) m X iff Gl(S) f X. Proof. Let G = Gl(S). ( ) By the theorem, any state s G(U) has a G-maximal extension s G(U ), for which U is successful by hypothesis, hence s has a successful extension. ( ) Any G-maximal s G(U) admits by hypothesis a successful extension which may only be on U by G-maximality, and hence U is successful. Tom Hirschowitz and Damien Pous Fair testing vs. must testing in a fair setting 36/38

Full abstraction for fair testing in CCS Tom Hirschowitz CNRS Abstract. In previous work with Pous [9, 8] (HP), we defined a semantics for CCS which may both be viewed as an innocent presheaf semantics

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